Chapter 6, Problem 15CR

(a) Find a polynomial function of degree $3$ whose y-intercept is $5$ and whose x-intercepts are $-2$, $3$ and $5$. Graph the function.

(b) Find a rational function whose y-intercept $5$ and whose x-intercepts are $-2$, $3$ and $5$ that has the line $x=2$ as a vertical asymptote. Graph the function.

To determine

The polynomial functions from function has degree $3$, $y-$ intercept is $5$ and $x$-intercepts are $-2,3,\text{and }5$ and graph the polynomial function.

Answer to Problem 15CR

Solution:

The equation of the graph is $\underset{\_}{y=\frac{x^3}{6}-x^2-\frac{x}{6}+5}$ and the graph is

Explanation of Solution

Given information:

The polynomial function has degree $3$, $y-$ intercept is $5$ and $x$-intercepts are $-2,3,\text{and }5$.

Explanation:

The $x$-intercepts $-2,3,\text{and }5$ are nothing but the real zeros of the polynomial function.

Factor theorem is $x=a$ is a zero of polynomial function $f\left(x\right)$, if and only if $\left(x-a\right)$ is a factor of $f\left(x\right)$.

As $x=-2$ is the real zero of polynomial function, corresponding is $\left(x-\left(-2\right)\right)=\left(x+2\right)$

For real zero $x=3$, corresponding factor is $\left(x-3\right)$ and for real zero $x=5$ corresponding factor is $\left(x-5\right)$

The required polynomial function would be multiplication of all these three factors.

The required polynomial function would be

$y=a\left(x+2\right)\left(x-3\right)\left(x-5\right)$, where $a$ is the leading coefficient to be determined.

It is given that $\left(x,y\right)=\left(0,5\right)$ is $y$-intercept of polynomial.

Substitute $x=0,y=5\text{in }y=a\left(x+2\right)\left(x-3\right)\left(x-5\right)$.

$\Rightarrow 5=a\left(0+2\right)\left(0-3\right)\left(0-5\right)$

$\Rightarrow 5=a\left(2\right)\left(-3\right)\left(-5\right)$

$\Rightarrow 5=30a$

Divide both sides by $30$

$\Rightarrow a=\frac{5}{30}$

Reduce the fraction to the simplestform.

$\Rightarrow a=\frac{1}{6}$

Substitute $a=\frac{1}{6}$ in $y=a\left(x+2\right)\left(x-3\right)\left(x-5\right)$

$y=\frac{1}{6}\left(x+2\right)\left(x-3\right)\left(x-5\right)$

Using distributive property,

$y=\frac{1}{6}\left(x+2\right)\left(x^2-5x-3x+15\right)$

$\Rightarrow y=\frac{1}{6}\left(x+2\right)\left(x^2-8x+15\right)$

$\Rightarrow y=\frac{1}{6}\left(x^3-8x^2+15x+2x^2-16x+30\right)$

$\Rightarrow y=\frac{1}{6}\left(x^3-6x^2-x+30\right)$

$\Rightarrow y=\frac{x^3}{6}-\frac{6x^2}{6}-\frac{x}{6}+\frac{30}{6}$

$\Rightarrow y=\frac{x^3}{6}-x^2-\frac{x}{6}+5$

Therefore, the required polynomial function is $y=\frac{x^3}{6}-x^2-\frac{x}{6}+5$

To graph the given polynomial function, plot the $y$-intercept $\left(0,5\right)$ and $x$-intercepts $x=-2,3,5$ on $xy$ coordinate system. Then, join all these points by a smooth curve.

Therefore, the graph of the polynomial $y=\frac{x^3}{6}-x^2-\frac{x}{6}+5$ is

To determine

The rational function from $y-$ intercept $5$, $x$-intercepts $-2,3,\text{and }5$ and vertical asymptote is line $x=2$, and then graph the rational function.

Answer to Problem 15CR

Solution:

The equation of the graph is $\underset{\_}{y=\frac{\left(x+2\right)\left(x-3\right)\left(x-5\right)}{3\left(2-x\right)}}$ and the graph is

Explanation of Solution

Given information:

The rational function has $y-$ intercept $5$, $x$-intercepts $-2,3,\text{and }5$ and vertical asymptote is line $x=2$.

Explanation:

The $x$-intercepts $-2,3,\text{and }5$ are nothing but the real zeros of the rational function.

So, corresponding factors $\left(x+2\right),\left(x-3\right),\left(x-5\right)$ would be in the numerator of rational function.

As line $x=2$ is a vertical asymptote, factor $\left(x-2\right)$ would be in the denominator of the rational function.

The required rational function would be

$y=\frac{a\left(x+2\right)\left(x-3\right)\left(x-5\right)}{\left(x-2\right)}$, where $a$ is the coefficient to be determined.

It is given that $\left(x,y\right)=\left(0,5\right)$ is $y$-intercept of polynomial

Substitute $x=0,y=5\text{in }y=\frac{a\left(x+2\right)\left(x-3\right)\left(x-5\right)}{\left(x-2\right)}$.

$\Rightarrow 5=\frac{a\left(0+2\right)\left(0-3\right)\left(0-5\right)}{\left(0-2\right)}$

$\Rightarrow 5=\frac{a\left(2\right)\left(-3\right)\left(-5\right)}{\left(-2\right)}$

$\Rightarrow 5=\frac{30a}{-2}$

$\Rightarrow 5=-15a$

Divide both sides by $-15$

$\Rightarrow a=\frac{5}{-15}$

Reduce the fraction to the simplest form

$\Rightarrow a=-\frac{1}{3}$

Substitute $a=-\frac{1}{3}$ in $y=\frac{a\left(x+2\right)\left(x-3\right)\left(x-5\right)}{\left(x-2\right)}$

$y=\frac{-\frac{1}{3}\left(x+2\right)\left(x-3\right)\left(x-5\right)}{\left(x-2\right)}$

$y=\frac{1\left(x+2\right)\left(x-3\right)\left(x-5\right)}{-3\left(x-2\right)}$

Using distributive property,

$\Rightarrow y=\frac{\left(x+2\right)\left(x-3\right)\left(x-5\right)}{3\left(2-x\right)}$

Therefore, the required rational function is $y=\frac{\left(x+2\right)\left(x-3\right)\left(x-5\right)}{3\left(2-x\right)}$

To graph the given rational function, plot the $y$-intercept $\left(0,5\right)$ and $x$-intercepts $x=-2,3,5$ on $xy$ coordinate system. Then, join all these points by a smooth curve having into consideration of vertical asymptote $x=2$

Therefore, the graph of rational function $y=\frac{\left(x+2\right)\left(x-3\right)\left(x-5\right)}{3\left(2-x\right)}$ is